Calculus of a Single Variable, Chapter 8, 8.2, Section 8.2, Problem 12

Recall that indefinite integral follows int f(x) dx = F(x) +C where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration.
For the given integral problem: int5 x/e^(2x) dx , we may apply Law of exponent: 1/x^n =x^(-n) then 1/e^(2x) = e^(-2x) .
int5 x/e^(2x) dx =int5 x*e^(-2x) dx
Apply the basic integration property: int c*f(x) dx = c int f(x) dx .
int5 x*e^(-2x) dx= 5int x*e^(-2x) dx
Apply integration by parts: int f *g' = f*g - int g *f'du .
Let : f =x then f' = dx
g'=e^(-2x) dx then g = -1/2e^(-2x)
Note: g = int g'= int e^(-2x) dx . Apply u-substitution using u =-2x then du = -2dx or (du)/(-2) =dx .
int e^(-2x) dx =int e^(u) * (du)/(-2)
= -1/2 int e^u du
= -1/2 e^u
Plug-in u =-2x on -1/2 e^u , we get: int e^(-2x) dx =-1/2 e^(-2x) .

Following the formula for integration by parts, we set it up as:
5int x*e^(-2x) dx =5[ x*(-1/2 e^(-2x)) - int (-1/2 e^(-2x)) dx]
=5[ x*(-1/2 e^(-2x)) - (-1/2) int ( e^(-2x)) dx]
=(-5xe^(-2x))/2 + 5/2int ( e^(-2x)) dx
Plug-in int e^(-2x) dx =-1/2 e^(-2x) , we get:
5int x*e^(-2x) dx=(-5xe^(-2x))/2 + 5/2int ( e^(-2x)) dx
=(-5xe^(-2x))/2 + 5/2*[-1/2 e^(-2x)] +C
=(-5xe^(-2x))/2 - (5 e^(-2x))/4 +C
or (-5x)/(2e^(2x)) - 5/(4 e^(2x)) +C